# History of Mathematics: Sophie Germain and Fermat's Last Theorem

Sophie Germain's greatest contribution to mathematics was in number theory. She discovered a special case of Fermat's Last Theorem which we now call the Germain Theorem.

Stated precisely: The equation $x^p + y^p = z^p$ has no non-zero integer solutions where $p$ is a Germain prime ($p$ is a prime number if $2p+1$ is also prime) and $p$ does not divide $xyz$. Could someone prove this? I can't find an actual proof of this statement or Germain's surviving works in English translation.

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This is proved in several books that treat Fermat's Last Theorem. But a really nice reference for this result (and much more) is Kenneth Ireland and Michael Rosen's beautuful book A Classical Introduction To Modern Number Theory. There the theorem is proved in just about a page in Chapter 17, section 4, which is actually entitled "Sophie Germain's Theorem". –  Adrián Barquero Apr 14 '12 at 19:08
Thank you for the reference I will look for the book –  Word Problems Apr 14 '12 at 19:18
The actual proof is given in the book's preview. Click the button and scroll down to page $275$. :) –  000 Apr 14 '12 at 19:19
See "Voici ce que j'ai trouvé:'' Sophie Germain's grand plan to prove Fermat's last theorem" by Laubenbacher and Pengelley, as well as "Unpublished manuscripts of Sophie Germain and a revaluation of her work on Fermat's last theorem" by A. del Centina. –  franz lemmermeyer May 5 '12 at 5:32

Sophie Germain's approach to the first case of Fermat's Last Theorem can be found in several textbooks that treat Fermat's Last Theorem. For example a very nice reference for her theorem is Kenneth Ireland and Michael Rosen's beautiful book A Classical Introduction to Modern Number Theory. There the theorem is proved in just about a page in Chapter 17, section 4, which is actually entitled "Sophie Germain's Theorem".

To add something to my comment let me mention a generalization of Sophie Germain's theorem which was found by Ernst Wendt.

Theorem (Wendt) Let $p$ be an odd prime. If there exists an integer $k \geq 1$ such that the number $q := kp + 1$ is also prime and satisfies

$$q \not \mid (k^k - 1)\operatorname{Res}{(X^k - 1, (X + 1)^k - 1)}$$

then the first case of Fermat's Last Theorem is true for the exponent $p$, that is, there are no solutions to $x^p + y^p = z^p$ with $p \not \mid xyz$.

Here the notation $\operatorname{Res}{(f, g)}$ stands for the resultant of the polynomials $f$ and $g$.

From this theorem we can obtain Sophie Germain's theorem as an inmediate corollary because in that case we are basically dealing with the case in which $k = 2$.

So we assume that $p$ is an odd prime such that $q = 2p + 1$ is also a prime number. Then we want to show that

$$q \not \mid (2^2 - 1)\operatorname{Res}{(X^2 - 1, (X + 1)^2 - 1)}$$

Now since $p > 2$ then $q > 5$ so certainly $q \not \mid (2^2 - 1)$. Also

$$\operatorname{Res}{(X^2 - 1, (X + 1)^2 - 1)} = \operatorname{Res}{(X^2 - 1, X^2 + 2X)} =$$

\begin{vmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ \end{vmatrix} $$= -3$$

so $q \not \mid \operatorname{Res}{(X^2 - 1, (X + 1)^2 - 1)}$ either. Then by Wendt's theorem we can conclude that the first case of FLT is true for the prime $p$, and thus we have Sophie Germain's theorem as a corollary of Wendt's result.

Note A reference for this result is Henri Cohen's book Number Theory Volume I: Tools and Diophantine Equations, it appears in section 6.9.4, page 430. Also observe that I used the determinant of the Sylvester matrix of the polynomials $X^2 - 1$ and $X^2 + 2X$ to compute their resultant.

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I've added added page numbers to the Google Books links in your post - I guess this might be useful for some users. (Of course, I am aware that they will not be displayed to everyone.) I hope you don't mind. –  Martin Sleziak May 5 '12 at 5:06
@MartinSleziak Thank you for adding the page numbers. I didn't know how to do that so that's why I only linked to the front cover of the books. –  Adrián Barquero May 5 '12 at 5:54